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Homework Help: Simple poles and residues

  1. Jan 8, 2009 #1
    I have a function 1/(z^4 + 1)

    I know the poles are pi/4, 3pi/4, 5pi/4, 7pi/4

    It says they are simple poles, I thought I understood why, but now I am totally confused. How does my lecturer just know that they are simple?

    A simple pole is a pole of order 1, but I thought this meant you look at the series and get the order from the highest indice of z on the denominator?

    Help!

    Cheers
     
  2. jcsd
  3. Jan 8, 2009 #2
    The numerator has simple zeroes, so the function is pof the form:

    1/[(z-a)(z-b)(z-c)(z-d)]

    where a, b, c, and d are different. If you expand this around, say, z = a, you can proceed as follows. You can write the function as:

    1/(z-a) * 1/[(z-b)(z-c)(z-d)]

    The factor 1/[(z-b)(z-c)(z-d)] is not singular at z = a, so it has a regular Taylor expansion. So, the Laurent expansion of the function around z = a is given by the Taylor expansion of that factor times 1/(z-a).
     
  4. Jan 8, 2009 #3
    What is a simple zero?
     
  5. Jan 8, 2009 #4

    HallsofIvy

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    Then you know wrong. The poles are [itex](\sqrt{2}/2)(1+ i)[/itex], [itex]-(\sqrt{2}/2)(1- i)[/itex], [itex](-\sqrt{2}/2)(1+ i)[/itex], and [itex](\sqrt{2}/2)(1- i)[/itex], which can be written [itex]e^{i\pi/4}[/itex], [itex]e^{3i\pi/4}[/itex], [itex]e^{5i\pi/4}[/itex], and [itex]e^{7i\pi/4}[/itex].

    [tex]\frac{1}{z^4+ 1}= \frac{1}{z- (\sqrt{2}/2)(1+ i)}\frac{1}{z+ (\sqrt{2}/2)(1-i)}\frac{1}{z+(\sqrt{2}/2)(1+ i)}\frac{1}{z-(\sqrt{2}/2)(1-i)}[/tex]
    Each of those numbers is a "simple pole" or "pole of order 1" because each has a power of 1 in the denominator.
     
  6. Jan 8, 2009 #5
    Cheers, this helps!
     
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